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Derivation of Velocity-Distance Gravity Equations
by Ron Kurtus (30 August 2009)
You can derive the gravity equations for the distance an object that is dropped, thrown downward or projected upward travels from the starting point when it reaches a given velocity, as well as the velocity for a given distance.
The equations t = (v − vi)/g and x = gt2/2 + vit, established in Derivation of Velocity-Time Gravity Equations and Derivation of Distance-Time Gravity Equations respectively, are used to determine the distance from the starting point when the object reaches a given velocity. The resulting equation leads to the relationship of velocity for a given distance.
The initial velocity of the object determines the starting direction. If the object is thrown upward, there are three possible values for distance or velocity, depending on the object's position.
Advanced Physics and Physical Science students are often required to derive equations. Beginning students who have not yet taken Calculus can skip or skim this lesson.
(See Overview of Derivation of Gravity Equations for a summary of the derivations.)
Questions you may have include:
- What is the distance for a given velocity equation?
- What is the velocity for a given distance equation?
- What is the effect of the initial velocity?
This lesson will answer those questions. There is a mini-quiz near the end of the lesson.
Useful tools: Metric-English Conversion | Scientific Calculator.
Distance for a given velocity
To determine the distance the object travels to reach a given velocity, start with the equations t = (v − vi)/g and x = gt2/2 + vit.
(These equations were obtained from Derivation of Velocity-Time Gravity Equations and Derivation of Distance-Time Gravity Equations respectively.)
Square both sides of t = (v − vi)/g:
t2 = (v − vi)2/g2
t2 = (v2 − 2vvi + vi2)/g2
Substitute the above equations for t2 and t into x = gt2/2 + vit:
x = g (v2 − 2vvi + vi2)/2g2 + vi(v − vi)/g
Multiply vi(v − vi)/g by 2/2 and combine like terms:
x = (v2 − 2vvi + vi2)/2g + 2(vvi − vi2)/2g
x = (v2 − 2vvi + vi2 + 2vvi − 2vi2)/2g
x = (v2 − vi2)/2g
Velocity for a given distance
To get the velocity for a given distance, multiply both sides of x = (v2 − vi2)/2g by 2g and solve for v:
2gx = (v2 − vi2)
v2 = 2gx + vi2
Take the square root of both sides of the equation:
v = ±√(2gx + vi2)
where
- ± means plus or minus
- √(2gx + vi2) is the square root of the quantity (2gx + vi2)
Note that squaring either +√(2gx + vi2) or −√(2gx + vi2) results in 2gx + vi2. Thus, ±√(2gx + vi2) is appropriate.
Also, plus-or-minus means there can be two solutions, one positive and one negative.
Effect of initial velocity
The direction of the initial velocity of the object influences velocity for a given distance equation, concerning both the plus-or-minus sign and the square root.
Since the direction of the force of gravity is downward, down is considered positive (+) and the direction up—away from the ground—is considered negative (−).
Thrown downward
When the object in thrown downward, vi, x and v are all positive numbers.
Distance
Thus, the equation of distance for a given velocity, x = (v2 − vi2)/2g, is unaffected. When the object is thrown downward, the velocity v is always greater than the initial velocity vi.
Velocity
The equation of velocity for a given distance is positive, since v is positive, and the following equation is used:
v = √(2gx + vi2)
Projected upward
When the object is thrown upward, it reaches some maximum height and then falls downward, going past the starting point.
The initial velocity vi is negative. The velocity v is negative when going up and positive when going down. The distance x is measured from the starting point. It is negative when going up and positive when going down.
When above starting point
Since the object moves upward and then falls downward, there are two solutions to the equation above the starting point, and the ± sign holds:
v = ±√(2gx + vi2)
Since x is negative when the object is moving upward, vi2 must be greater than 2gx, because taking a square root of a negative number is not permitted.
If x is negative and 2gx = vi2, then v = 0 and the object is at its maximum height.
When x = 0:
v = ±√(vi2)
v = ±vi2
That means that v is moving up at x = 0 and also down at x = 0.
When below starting point
After the object has passed below the starting point, x and v are positive numbers. The positive version of the velocity equation holds:
v = √(2gx + vi2)
Initial velocity is zero
In the situation where the object is simply falling, vi = 0 and the equations become:
x = v2/2g
v = √(2gx)
Summary
You can derive the gravity equations for the distance an object that is dropped, thrown downward or projected upward travels from the starting point when it reaches a given velocity, as well as the velocity for a given distance.
The equations t = (v − vi)/g and x = gt2/2 + vit, are used to determine the distance for a given velocity. The resulting equation leads to the relationship of velocity for a given distance.
The initial velocity of the object determines the starting direction. If the object is thrown upward, there are three possible values for distance or velocity, depending on the object's position.
The derived equations are:
x = (v2 − vi2)/2g
v = ±√(2gx + vi2) above the starting point
v = √(2gx + vi2) below the starting point
See the Side Menu for more Gravitation and Gravity topics
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Resources
The following resources provide information on this subject:
Websites
Acceleration due to Gravity Calculations - from Western Washington University
Gravitation and Gravity Resources
Books
Top-rated
books on Simple Gravity Science
Top-rated
books on Advanced Gravity Physics
Mini-quiz to check your understanding
If you got all three correct, you are on your way to becoming a Champion in Physics. If you had problems, you had better look over the material again.
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